Problem: $f(x, y) = \tan(x)y^{2.5}$ $\dfrac{\partial^3 f}{\partial y^3} = $
Solution: Taking a third order partial derivative is like taking a regular third order derivative. We take the partial derivative once, then we take another partial derivative, and then we take one more. $\dfrac{\partial^3 f}{\partial y^3} = \dfrac{\partial}{\partial y} \left[ \dfrac{\partial}{\partial y} \left[ \dfrac{\partial f}{\partial y} \right] \right]$ Let's differentiate! $\begin{aligned} \dfrac{\partial^3 f}{\partial y^3} &= \dfrac{\partial}{\partial y} \left[ \dfrac{\partial}{\partial y} \left[ \dfrac{\partial}{\partial y} \left[ \tan(x)y^{2.5} \right] \right] \right] \\ \\ &= \tan(x) \dfrac{\partial}{\partial y} \left[ \dfrac{\partial}{\partial y} \left[ \dfrac{5}{2} y^{1.5} \right] \right] \\ \\ &= \dfrac{5}{2} \tan(x) \dfrac{\partial}{\partial y} \left[ \dfrac{3}{2} y^{0.5} \right] \\ \\ &= \dfrac{15}{4} \tan(x) \dfrac{y^{-0.5}}{2} \\ \\ &= \dfrac{15 \tan(x)}{8 \sqrt{y}} \end{aligned}$ Therefore, $\dfrac{\partial^3 f}{\partial y^3} = \dfrac{15 \tan(x)}{8 \sqrt{y}}$.